Bourbaki

Exercise 6, § 7

Algebra, III, 634

Bourbaki marks this one as harder than the rest.

Let K be a commutative field and E a vector space over K.

(a) For an element z=∑p=0∞zpz = \sum_{p=0}^\infty z_p (with zp∈⋀pEz_p \in \bigwedge^p E) of the exterior algebra ⋀E\bigwedge E of E to be invertible, it is necessary and sufficient that z0≠0z_0 \neq 0 (prove it first when E is finite-dimensional and then pass to the general case, noting that for all z∈⋀Ez \in \bigwedge E there exists a finite-dimensional subspace F of E such that z∈⋀Fz \in \bigwedge F).

(b) Suppose that K is such that 2≠02 \neq 0 in K. Show that, if E is infinite-dimensional or of even finite dimension, the centre of the algebra ⋀E\bigwedge E consists of the elements such that zp=0z_p = 0 for all odd indices pp. If E is of odd dimension nn, the centre of ⋀E\bigwedge E is the sum of the above subspace and ⋀nE\bigwedge^n E. In all cases, the centre of ∧E\wedge E is identical with the set of elements of ∧E\wedge E permutable with all the invertible elements of ∧E\wedge E.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.